Question

# The heat liberated on complete combustion of $$7.8 \,g$$ benzene is $$327 \,kJ$$. This heat was measured at constant volume and at $$27^oC$$. Calculate the heat of combustion of benzene at constant pressure.

A
3274 kJmol1
B
1637 kJmol1
C
3270 kJmol1
D
None of the above

Solution

## The correct option is B $$-3274\ kJ mol^{-1}$$Solution:- (A) $$- 3274 \; {kJ}/{mol}$$Molecular weight of $${C}_{6}{H}_{6} = 78 \; g$$Heat liberated on combustion of $$7.8 \; g$$ benzene $$= 327 \; kJ$$Heat liberated on combustion of $$78 \; g$$ benzene $$= \cfrac{327}{7.8} \times 78 = 3270 \; kJ$$$$\therefore \Delta{E} = -3270\ kJ$$Here negative sign indicates that the energy is liberated.$${{C}_{6}{H}_{6}}_{\left( l \right)} + 7 \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow 6 {C{O}_{2}}_{\left( g \right)} + 3 {{H}_{2}O}_{\left( l \right)}$$From the above reaction,$$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = 6 - \cfrac{15}{2} = - \cfrac{3}{2}$$Temperature $$\left( T \right) = 25 ℃ = \left( 25 + 273 \right) K = 298 K \; \left( \text{Given} \right)$$Now from first law of thermodynamics,$$\Delta{H} = \Delta{E} + \Delta{{n}_{g}} RT$$$$\Delta{H} = \Delta{E} + \left( -\cfrac{3}{2} \right) \times 8.314 \times {10}^{-3} \times 300$$$$\Rightarrow \Delta{H} = - 3270 - 3.74 = - 3273.74 \; {kJ}/{mol} \approx -3274 \; {kJ}/{mol}$$Hence the heat of reaction at constant pressure will be $$- 3274 \; {kJ}/{mol}$$.Chemistry

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